A and B decompose via first order kinetics with half-lives 54.0 min and 18.0 min respectively. Starting from an equimolar non-reactive mixture of A and B, the time taken for the concentration of A to become 16 times that of B is ________ min. (Round off to the Nearest Integer).

Both A and B follow first-order kinetics with half-lives $$t_{1/2}^A = 54.0$$ min and $$t_{1/2}^B = 18.0$$ min respectively. Starting from equimolar concentrations, let the initial concentration of each be $$C_0$$.

At time $$t$$, the concentrations are: $$[A] = C_0 \left(\frac{1}{2}\right)^{t/54}$$ and $$[B] = C_0 \left(\frac{1}{2}\right)^{t/18}$$.

We need $$[A] = 16[B]$$: $$C_0 \left(\frac{1}{2}\right)^{t/54} = 16 \cdot C_0 \left(\frac{1}{2}\right)^{t/18}$$.

Dividing both sides by $$C_0$$ and rearranging: $$\left(\frac{1}{2}\right)^{t/54} = 2^4 \cdot \left(\frac{1}{2}\right)^{t/18}$$.

Rewriting $$2^4 = \left(\frac{1}{2}\right)^{-4}$$: $$\left(\frac{1}{2}\right)^{t/54} = \left(\frac{1}{2}\right)^{t/18 - 4}$$.

Equating exponents: $$\frac{t}{54} = \frac{t}{18} - 4$$. This gives $$\frac{t}{54} - \frac{t}{18} = -4$$, so $$\frac{t - 3t}{54} = -4$$, hence $$\frac{-2t}{54} = -4$$.

Solving: $$t = \frac{4 \times 54}{2} = 108$$ min.

The time taken is $$108$$ min.